Ethanol and methyl cyclohexane have vapour pressures of 22.4 kPa and 37.3 kPa respectively at 328 K. A mixture of the two in which the mole fraction of methyl cyclohexane is 0.32 has total vapour pressure of 50.1 kPa at the same temperature. These solvent mixtures

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Ethanol and methyl cyclohexane have vapour pressures of 22.4 kPa and 37.3 kPa respectively at 328 K. A mixture of the two in which the mole fraction of methyl cyclohexane is 0.32 has total vapour pressure of 50.1 kPa at the same temperature. These solvent mixtures

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show negative deviations from Raoult's law

show positive deviations from Raoult's law

do not form azeotropes

are ideal mixtures

answer is D.

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Detailed Solution

Total vapour pressure as calculated from Raoult's law
$= p = p_{B}^{\circ} + x_{A} (p_{A}^{\circ} - p_{B}^{\circ})$
or $p = 22 .4 + 0 .32 (37 .3 - 22 .4) = 27 .2 kPa$
since the actual vapour pressure is higher than that predicted by Raoult's law, these mixtures show positive deviations from Raoult's law